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Sindrei [870]
2 years ago
8

Polarizing windows, filters, etc. are often used to reduce the amount of light that enters the lens of a camera or into a room o

r a car. A library atrium has an overhead skylight that lets in too much light during the day which heats up the interior of the library far too much. The building engineer installs new double paned polarizing sky lights to reduce the intensity. If sunlight, which is unpolarized, has an average intensity of 1250 W/m^2.
Required:
What angle should the polarizing axis of the second pane of the window make with the polarizing axis of the first pane of the window in order to reduce the intensity of the sunlight to 33% of the original value?
Physics
1 answer:
LekaFEV [45]2 years ago
7 0

Answer:

The answer is "35.6^{\circ}"

Explanation:

The sunlight level of the first panel:

I_1 = \frac{I_o}{2}

When the light of this intensity passes through the second window:

I_2 = I_1 \cos^2 \theta\\\\I_2 = \frac{I_o}{2} \cos^2 \theta

 \frac{I_2}{I_o} = 0.33 (33\%) \\\\

therefore,  

0.33 = \frac{1}{2} \cos^2 \theta\\\\\cos^2 \theta = 0.66\\\\\cos \theta = \sqrt{0.66} = 0.8124\\\\\theta = \cos^{-1}( 0.8124) = 35.6^{\circ}\\\\

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Select the correct answer.
Pachacha [2.7K]

Answer:

c

Explanation:

It's c the last one u see

8 0
2 years ago
Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the fre
liq [111]

Answer:  D. the wave has traveled 97.2 cm in 1 second.

Explanation:

Got it right on edge, hope this helps :)

7 0
3 years ago
Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 4.0 m behind the apertur
devlian [24]

Answer:

d = 0.247 mm

Explanation:

given,

λ = 633 nm

distance from the hole to the screen = L = 4 m

width of the central maximum = 2.5 cm

                                             2 y = 0.025 m

                                               y = 0.0125 m

For circular aperture

  sin \theta = 1.22\dfrac{\lambda}{d}

using small angle approximation

  \theta = \dfrac{y}{D}

now,

   \dfrac{y}{D} = 1.22\dfrac{\lambda}{d}

   y = 1.22\dfrac{\lambda\ D}{d}

   d = 1.22\dfrac{\lambda\ D}{y}

   d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}

         d =0.247 x 10⁻³ m

         d = 0.247 mm

the diameter of the hole is equal to 0.247 mm

5 0
3 years ago
Which of the statements best describes density?
Vesnalui [34]

Answer:

The answer to your question is: letter D.

Explanation:

a.The mass that a mole of substance has, measured in grams per mole.  Density is not measure in moles, so this is not the correct answer.

b.The amount of substance dissolved in a liquid, measured in moles per liter.  The substance dissolved in a liquid must be measure in grams not in moles, so this answer is incorrect.

c.The mass of substance dissolved in a liquid, measured in grams per milliliter.  I think that this definition is correct but is incomple, so this answer is wrong.

d.The ratio of a substance's mass to its volume, measured in grams per milliliter and also equivalent to grams per cubic centimeter. This is the right description to density, so this is the correct answer.

8 0
3 years ago
A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800
Arada [10]

Explanation:

The given data is as follows.

            m = 5000 kg,            h = 800 km = 0.8 \times 10^{6} m

    R_{e} = 6.37 \times 10^{6} m,    r = R + h = 7.17 \times 10^{6} m

   M_{e} = 5.98 \times 10^{24} kg,   G = 6.67 \times 10^{-11} Nm^{2}/kg^{2}

As we know that,

              \frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}

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And, it is known that formula to calculate angular velocity is as follows.

               \omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}

                            v = \sqrt{\frac{GM_{e}}{r^{3}}}

                               = \sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}

                                = 1.0402 \times 10^{-3} rad/s

Thus, we can conclude that speed of the satellite is 1.0402 \times 10^{-3} rad/s.

6 0
3 years ago
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